Jeans Instability and

Gravity introduces a new element into the physics of waves (aka stable perturbations) in a gas. Simply put, a gas cloud’s self gravity can cause perturbations on sufficiently large scales to become unstable. The goal here is to present a brief overview of fluid dynamics, and to give both rigorous and non-rigorous derivations of the Jeans criterion (named after Sir James Jeans, who first published this analysis in 1902) for gravitational instability. We set the stage by first considering sound waves in a non-gravitating gas, establishing some basic methodology and results. Then we include gravity, and ask how it changes the picture.

Fluid Dynamics 101

We consider a fluid in which quantities such as the number and mass n(x, t) and ρ(x, t), the P (x, t), the T (x, t), and the velocity field v(x, t) can all sensibly be defined at any point x. As a practical matter, this means that the fluid must be such that the local interparticle spacing ` = n−1/3 is much less than the scale of interest λ. Many (but not all) astrophysical systems fall into this category. The behavior of the fluid is governed by three fundamental equations. First, the continuity equation is a statement of conservation of mass at any point: ∂ρ + ∇ · (ρv) = 0. (1) ∂t Integrating this equation over any small volume V surrounding the point x and applying the R divergence theorem to the second term, this equation just says that the rate of change of V ρdV , H the total mass within V , is equal to S ρv · .dS, the rate at which mass flows outward across the surface of V . The second fundamental equation is the Euler equation, which is just the fluid equivalent of Newton’s second law: ∂v ∇P + v · ∇v = − . (2) ∂t ρ The right-hand side is just the pressure gradient, the force per unit volume on the fluid, divided by the , the mass per unit volume. The left-hand side is the acceleration of the fluid. We can understand the latter statement by considering the rate of change of any quantity Q(x, t) that describes the fluid. As we move with a fluid element, Q changes for two reasons. First, the value of Q at any point x is changing if ∂Q/∂t 6= 0 there. Second, as the fluid moves, the element samples different regions of space, which will have different values of Q if ∇Q is nonzero. Applying the chain rule for partial differentiation, we have, writing x = (x, y, z), ∂Q ∂Q ∂Q ∂Q Q(x + δx, y + δy, z + δz, t + δt) − Q(x, y, z, t) ≈ δx + δy + δz + δt, ∂x ∂y ∂z ∂t so dQ Q(x + δx, y + δy, z + δz, t + δt) − Q(x, y, z, t) ≡ lim dt δ→0 δt ∂Q ∂Q ∂Q ∂Q = v + v + v + ∂x x ∂y y ∂z z ∂t ∂Q = v · ∇Q + , ∂t

1 where v = (vx, vy, vz) = limδt→0 δx/δt. The quantity dQ/dt, the rate of change of Q moving with the fluid, is sometimes called the convective derivative of Q. Setting Q = v, we see that the right hand side of Equation (2) is simply dv/dt. Note that the right-hand side of Equation (2) as written contains only the pressure force. Other forces, such as magnetic fields or internal viscous terms, could also appear there, but we will neglect them here. Equations (1) and (2) represent 4 equations (1 scalar, 1 vector) in 5 unknowns (ρ, P , and v). We need another equation to find a solution. That equation is the of the fluid

P = P (ρ, S) (3) where S is entropy. We will confine ourselves here to isentropic flows, with S = constant. These equations have many interesting solutions, but our interest here is more limited. To start, let’s consider a uniform fluid and look at the behavior of small perturbations to that background solution. Specifically, the unperturbed fluid motion is steady, ∂/∂t ≡ 0, with uniform density and pressure ρ0 and P0, and zero velocity v0 = 0. You can easily verify that this solution trivially satisfies Equations (1) and (2). Now let’s consider a small perturbation to this baseline solution. We write

ρ = ρ0 + ρ1

P = P0 + P1

v = v1 and substitute these into Equations (1) and (2): ∂ρ ∂ρ 0 + 1 + ∇ · [(ρ + ρ )v ] = 0 (4) ∂t ∂t 0 1 1 ∂v1 ∇(P0 + P1) + v1 · ∇v1 = − . (5) ∂t ρ0 + ρ1 We then linearize these equations by expanding them to first order, throwing away all products of small terms (having subscript 1). This is the basic mathematical technique employed in all studies of the stability of nonlinear dynamical systems—find an equilibrium solution, then look at the first-order equations describing small perturbations to that solution. (Why? Because we know how to solve linear equations and they give us valuable insight into the physics of the system.) In this case, discarding all products involving more than one small term, and imposing the properties of the unperturbed solution, Equations (4) and (5) become ∂ρ 1 + ρ ∇ · v = 0 (6) ∂t 0 1 ∂v ρ 1 = −∇P . (7) 0 ∂t 1 Note that the tricky nonlinear v · ∇v term in the Euler equation, which is the source of a host of complex physical phenomena in real fluids, ranging from turbulent dissipation to shock waves, has disappeared. The key point is that Equations (6) and (7) are soluble. Taking the partial derivative of Equation (6) with respect to time and taking the divergence of Equation (7), and noting that ∂(∇ · v1)/∂t = ∇ · (∂v1/∂t), we can eliminate the v1 terms and find

∂2ρ 1 − ∇2P = 0. (8) ∂t2 1

2 We now use the equation of state, Equation (3), to obtain

∂P ∂P P1 ≡ δP = δρ = ρ1 , ∂ρ S,0 ∂ρ S,0 so Equation (8) becomes 2 ∂ ρ1 ∂P 2 2 − ∇ ρ1 = 0. (9) ∂t ∂ρ S,0 We recognize Equation (9) as the wave equation, the solutions to which are traveling waves with wave speed cs, where

2 ∂P cs = . (10) ∂ρ S,0 For an with P = ρkT/m = Aργ, the adiabatic sound speed is s q γP c = Aγργ−1 = s ρ s γkT = m  T 1/2  m −1/2 = 0.91 km/s γ1/2 , (11) 100 K mH where m is the mean molecular weight. Linearity means that any linear combination of solutions is also a solution. In that case, since any solution can be decomposed into a Fourier series (for a finite region) or a Fourier transform (for an infinite domain), either of which is just a sum of plane-wave solutions of the form

ik·x−iωt ρ1 ∝ e (12)

(and similarly for other quantities), we can understand the physics of the system by considering the behavior of plane-wave modes. Here ω = 2πf is the angular frequency and k is the wave vector, whose direction is the direction of motion of the plane wave and whose magnitude (the wavenumber) is k = 2π/λ, where λ is the wavelength. Substituting Equation(12) into Equation (9) we obtain the familiar relation between wavenum- ber and frequency:

ω = cs k (13) or fλ = cs. (14)

Self-gravitating fluids

Now let’s consider how the above analysis changes when gravity is included. Gravity leaves the continuity equation (1) unchanged, and introduces an additional force into the right-hand side of the Euler equation (2), which now becomes

∂v ∇P + v · ∇v = − − ∇φ, (15) ∂t ρ

3 where the gravitational acceleration is a = −∇φ and the gravitational potential φ satisfies Poisson’s equation ∇2φ = 4πGρ. (16) In the spirit of the earlier approach, we seek a steady (∂/∂t = 0) unperturbed solution with uniform density ρ0 and pressure P0, and zero velocity. We note in passing that this formally leads to a contradiction for the potential: Substituting the “0” quantities into the Euler equation implies 2 ∇φ0 = 0, so φ0 = constant, but Poisson’s equation (16) says ∇ φ > 0. This contradiction is sometimes called the Jeans swindle. Binney & Tremaine (p. 402) discuss some arguments to get around it in practice; see also the article by Kiessling (2003, Adv.Appl.Math. 31, 132; arXiv:astro- ph/9910247v1). In any case, this technical problem with the unperturbed state is not regarded as a serious flaw in the perturbation analysis. We might imagine that other forces, such as rotation or magnetism, are also at play in the initial state, allowing a self-consistent solution to exist. For example, consider a homogeneous self-gravitating fluid of density ρ contained within a rotating cylinder of radius R. The cylinder and the fluid rotate at angular speed Ω about the axis of the cylinder. It is readily shown from Equations (15) and (16) that the fluid can be in true equilibrium, with no pressure gradient (i.e. the gravitational and centrifugal forces balance), if Ω2 = 2πGρ. Ignoring the details of the Jeans swindle and writing

ρ = ρ0 + ρ1

P = P0 + P1

v = v1

φ = φ0 + φ1, substituting into Equations (1), (15), and (16), and again using Equation (10) to eliminate P1, we obtain the linearized equations ∂ρ 1 + ρ ∇ · v = 0 (17) ∂t 0 1 ∂v ρ 1 = −c2∇ρ − ∇φ . (18) 0 ∂t s 1 1 2 ∇ φ1 = 4πGρ1 (19)

Combining, as before, the time derivative of Equation (17) with the divergence of Equation (18) 2 and employing Equation (19) to eliminate ∇ φ1, we arrive at

∂2ρ 1 − c2∇2ρ = − 4πGρ ρ , (20) ∂t2 s 1 0 1 which is identical to Equation (9) apart from the additional term on the right-hand side. Again looking for plane-wave solutions, we substitute Equation (12) into Equation (20) to obtain the new relation between wavenumber and frequency:

2 2 2 ω = cs k − 4πGρ0 . (21)

The difference between Equations (13) and (21) is profound. In the former case, ω2 is positive (and so ω is real) for any k. In the latter, since the second term on the right is negative, it is possible for ω2 to be negative for some k. What does this mean? Setting ω2 = −α2, we have ω = ±iα and eiωt = e∓αt. In other words, the oscillatory behavior found for real ω is replaced

4 by exponential growth for imaginary ω—the disturbance is unstable. Any wavenumber k leading to ω2 < 0 represents a perturbation whose amplitude grows exponentially in time. An overdense region becomes denser and denser, leading to gravitational collapse. From Equation (21), it is clear that all plane-wave perturbations having √ 4πGρ0 k < kJ ≡ (22) cs are unstable. The critical wavenumber kJ is called the Jeans wavenumber. In terms of wavelength λ = 2π/k, this means that all perturbations having r π λ > λJ ≡ cs (23) Gρ0 are unstable. The critical wavelength λJ is the Jeans length. All scales larger than λJ are gravitationally unstable. It is common to recast the above criteria in terms of the total mass involved. The mass contained within a spherical volume having a diameter equal to the Jeans length is called the Jeans mass, MJ , where 4π π5/2 c3 1 3 s MJ ≡ ρ0 ( λJ ) = (24) 3 2 6 3/2 1/2 G ρ0 Some Simpler Arguments

Armed with the exact solution, we can explore some simpler, “back of the envelope” arguments to understand the result. First, as discussed in Binney & Tremaine, let’s imagine compressing a spherical region of radius R in a homogeneous gas cloud. If R → (1 − α)R, where α is small and positive, and the mass 3 M remains constant, then ρ0 ∝ M/R → (1 + 3α)ρ0. From Equation (10), the pressure change is 2 3 δP = csδρ = 3αcsρ0 and so—to order of magnitude at least—the pressure gradient between the 3 center of the region and the edge is ∼ δP/R = 3αcsρ0/R and the (outward) pressure force per unit mass is 2 δP 3αcs ap ≈ ≈ . (25) ρ0R R As the region is compressed, the gravitational acceleration at its surface, GM/R2, also increases. The extra (inward) gravitational acceleration due to the compression is 2GMα 8πGρ Rα a ≈ = 0 . (26) g R2 3 4 3 where we have used the fact that M = 3 πR ρ0. Comparing Equations (25) and (26), we see that the gravitational force dominates (and continues to dominate) if ap < ag, or 9 c2 R2 > s . (27) 8π Gρ0 Apart from the numerical coefficient, we see that the critical value of R is just the Jeans length (Equation 23). Following Sparke & Gallagher (p. 355) we can reach the same conclusion by considering the thermal and potential energies of the region. The total gravitational potential energy of a homo- 4π 3 2 5 geneous sphere of radius R and mass M = 3 R ρ0 is easily shown to be U = −3GM /5R ∝ R ,

5 3 1 2 3 while the thermal energy (for an ideal gas) is K = 2 MkT/m = (3/γ) 2 Mcs ∝ R , from Equation (11). Comparing the two, we see that the total energy U + K becomes negative (i.e. the region’s gravity dominates its thermal energy) for

3 3 GM 2 Mc2 < 2γ s 5 R so 2γ GM 8πγ c2 < = Gρ R2 s 5 R 15 0 or 15 c2 R2 > s , (28) 8πγ Gρ0 again essentially the same result as Equations (23) and (27). Finally, we note that the instability criterion (Equation 23) can be rewritten (again neglecting the numerical factors) as

R > −1/2 ∼ (Gρ0) . (29) cs The right-hand side is the free-fall time for the clump—the time scale for it to collapse under gravity if pressure were negligible—while the left-hand side is the time taken for a sound wave to cross the system. The Jeans criterion can thus be interpreted as meaning that sound cannot traverse the region (and hence pressure cannot operate) in time to prevent the collapse. All of the above examples confirm that the Jeans instability is the result of competition between thermal (pressure/sound speed) forces and gravitational forces. In each case, however the gravity is quantified, gravitational collapse ensues when gravity dominates over the thermal term.

Consequences of the Jeans Instability

Let’s evaluate the Jeans length and mass, Equations (23) and (24), for parameters of astrophysical interest. Plugging in numbers typical of dense molecular cores (with particle mass m = 3.3 × 10−24 g), we obtain

 T 1/2  n −1/2 λ = 1.0 pc (30) J 10 K 103 cm−3  T 3/2  n −1/2 M = 26 M , (31) J 10 K 103 cm−3 where cs = 260 m/s for T = 10 K and γ = 5/3. Thus any dense molecular core containing more than a few tens of solar masses of gas is unstable, and will collapse in roughly a free-fall time

 n −1/2 t = (Gρ)−1/2 = 2.1 Myr . (32) ff 103 cm−3 As the unstable region collapses its density rises, but the atomic and molecular processes cooling it remain efficient so long as the cloud is optically thin to the cooling radiation, and the temperature remains roughly constant. As a result, the Jean’s mass steadily decreases as the collapse proceeds, and the collapsing cloud fragments into lower and lower mass pieces, each collapsing on its own free-fall time scale. Our initial collapsing cloud is on its way to becoming a cluster.

6 The process ends when the fragments become so dense that they are optically thick to the radiation cooling them, so the radiation can no longer escape and the temperature begins to rise, stabilizing the collapse and forming a . Experts and simulations disagree on precisely when this point is reached, and on the subsequent evolution of the , but this is the subject of active research. At stake is our understanding of the stellar mass function and the physical state of young star clusters.

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